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I found the easiest way to convince non-mathematical types is by suggesting we do the million-door version: “Ok, pick a door between 1 and 1,000,000. We’ll call
by drfuchs 7y ago
I found the easiest way to convince non-mathematical types is by suggesting we do the million-door version: “Ok, pick a door between 1 and 1,000,000. We’ll call that door N. Now, I’ll open 999,998 bad doors. So, now only door N and door 223,205 are still closed, and one of them has the good prize. Want to keep door N, or swap?”
- avn2109 7y agoGosh this makes it much easier.
- turc1656 7y agoThis is an excellent way to make it click for people.
- graeme 7y agoThis is exactly how I conceived of it when I first heard of the problem. Pretty sure I remember demonstrating it to people using 10+ playing cards and flipping over all but one of the cards the person didn’t choose.
- CydeWeys 7y agoThe flaw with this thought experiment as a way of teaching the original problem is that it's not at all the same to reveal 998,000 bad doors as it is to reveal just 1, and when revealing just 1 in this situation the effect is much smaller than just revealing 1 of 3 doors.
- deleted 7y ago[deleted]
- nwallin 7y agoPlease explain to me why it is different whether to switch doors after opening all but one door vs after opening all but one door? (note: when you do this in real life, you stop talking now. let them explain it.) When you have N>2 doors, the host will with 100% certainty open N-2 doors with goats behind them. It doesn't matter whether N is 3 or 1,000,000. After the host's hidden knowledge has reduced the problem space from N options to 2 options, the probability of getting the car is increased from 1/N to 1/2. The key phrase is "host's hidden knowledge". Once the host starts opening the wrong doors, the fundamental nature of the problem is changed from being an independent probability problem to a dependent probability one.
- henrikschroder 7y ago> After the host's hidden knowledge has reduced the problem space from N options to 2 options, the probability of getting the car is increased from 1/N to 1/2. No, your numbers are wrong. In your initial choice, the probability of picking the winning door is 1/N. When Monty opens doors so that there is a single door remaining, the probability of that door containing the prize is the complement of the probability that you picked the winning door, i.e. 1-(1/N). So switching doors increases your probability from 1/3 to 2/3 if N=3, not to 1/2. If N=1000, switching doors increases your probability from 1/1000 to 999/1000, and that is why increasing the number of doors makes the correct solution more intuitive for some people.
- CydeWeys 7y agoAnd the point I'm making is that it's not clear at all why opening 998 doors out of 1,000 is analogous to opening 1 door out of 3 -- why wouldn't you still be opening just the 1 door? If you pick one of 1,000 doors, and then the host reveals a goat behind a single door, there's a slight (but only very slight) advantage to switching. The advantage is most pronounced with the fewest possible number of doors, i.e. 3. That's why this approach to teaching it has never made sense to me -- it's not clear at all why opening 998 doors should be analogous to opening 1 door, which it's much more obvious that opening 1 door is analogous to 1 door.
- ivanb 7y agoAgain you didn't say if you opened 999998 bad doors due to chance or by conscious decision. If you did it by chance then the answer would be completely different.
- selestify 7y agoHow would the answer be any different? As long as the doors are bad doors, it doesn't matter if they're opened by chance or not
- mdonahoe 7y agoWeirdly, it matters. You can simulate it. I didn't believe it until I simulated it myself.
- tiziano88 7y agoHow did you count the runs in which the good door was accidentally opened?
- grraaaaahhh 7y agoThose runs get ignored since that isn't the scenario. Specifically we're looking at the case where Monty opens a goat by chance (ie. he could have opened the car, but did not). It also helps that that result doesn't matter to the overall question, is it better to switch or to stay, since both options are irrelevant when Monty opens the car door.
- semiotagonal 7y agoRESCINDED: Error in my simulation! Apologies.
- sparky_z 7y agoYou're accidentally including the cases where montysPick == myPick. You have to discard those cases. Monte has to show you what's behind a different door than the one you initially picked. The game doesn't make sense otherwise.
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- itcrowd 7y agoWhat made me convince friends is the following (using 3 cards, one red and two black): - Let them pick one card of three - Offer to switch their card for the other two (closed!) cards. Most people take this deal, in my experience everyone takes this deal. - Then, show the "goat"/useless card of the two remaining cards - Offer them (again) to switch their original choice of card for the remaining card. If they accept to switch, let them explain why (this is when people get it). If they don't switch, tell them that one of the two cards they could have switched for must have been a goat, so I could always show them one goat between these two. After this, most people (all my test subjects, anyway) understand that in the long run, they're best off switching to the "two card" deal. They now understand that one must be a goat but that the probability of the "car" being among the two-card offer is twice as likely. N<10 so YMMV
- v64 7y agoI've had similar experiences with my friends. The concept is hard to grasp in the abstract, but if you actually play it out and keep track of how often they win and lose, people are generally willing to accept the correctness of the result, even if they don't grasp the proof behind it.
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- mjevans 7y agoI think the fault is one of false temporal equivalency. When a casual observer assesses the situation they see an equal set of unknowns behind each outcome, since that matches the set of conditions. While the probability would have remained the same, it's the elimination of all (but one) remaining false choices that tips the scale. The weight that the selected door was the correct one remains at 1/3rd, but the probability that the only remaining door was the correct one is the inverse, 2/3rds. Meanwhile in the 1,000,000 doors set, if instead one door was picked, one other door was opened, and the choice was between keeping the current door and picking one of the other 999,998 doors, the odds increase so infinitesimally that both ego and a need to know would bias me to staying the course.
- cortesoft 7y agoMy method of explaining it is that you can think of it as getting to pick 1 of 3 doors or 2 of the 3 doors... your original pick has a 1/3 chance... 2/3 of the time it will be one of the other two... if it was not the door you picked, it HAS to be the one that you didn't pick and wasn't opened... meaning the non-picked, non-opened door has a 2/3 chance.
- majikandy 7y agoYes this is the exact explanation I give. Mentally pick 2 doors, but say out loud the other 1, when offered the switch you take it as it means you are back on your secretly pre-selected 2. Hence 2/3 chance by always switching.
- monkeydreams 7y ago> 2/3 of the time it will be one of the other two I was debugging my monty-hall simulator to try and work out why my intuition was so wrong when I came across my comment "Iterate through doors, open first door without prize" when this point finally struck me.
- jhncls 7y agoWith more than 3 doors, it is essentially a different problem. Imagine an evil host who only opens a door and offers to switch when the guest chose the door with the car, and not offer to switch during other episodes. Or, more evil, infrequently open a bad door and offer to switch, just to confuse the candidate. With 1,000,000 doors such an evil host will never open and offer the switch.
- sushid 7y agoIt’s really not. It goes from a 1/3 vs 1/2 question to a 1/1MM vs 1/2, which just makes the difference in odds more easy to perceive.
- henrikschroder 7y agoWhere do you get the 1/2 probability from?
- melling 7y agoLeonard Mlodinov uses 100 doors to illustrate this on page 55 of his book, which I enjoyed. https://en.m.wikipedia.org/wiki/The_Drunkard's_Walk https://en.m.wikipedia.org/wiki/The_Drunkard's_Walk
- leepowers 7y agoAlternatively, the way I think of it is as follows: An initial random guess has a 1 out of 3 chance of being correct. Being shown a goat doesn't change this probability. Staying with the first choice doesn't change this probability. Only making another choice, and choosing from the smaller pool of two doors increases the odds. Or perhaps better stated in the Wikipedia article: > "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"
- dllthomas 7y agoBut being shown a goat can change the probability. If Monty picks randomly (which is not the original problem but is interesting to consider) then there is no benefit to switching.
- rags2riches 7y agoMonty picking randomly adds a third outcome. You then have a) Staying wins (Monty opens to a goat) b) Swap wins (Monty opens to a goat) c) First pick lost, no chance to swap (Monty opens to the car) each of equal probability.
- leepowers 7y agoVery true! It depends on the particular variation[1] of the game being played. Which is one difficulty of MH - the ambiguity. Probabilities change when the initial conditions are slightly different or when the problem is interpreted differently. I've never assumed Monty picks at random. But it's entirely possible someone else would make that assumption, in turn changing the optimal strategy. [1] https://en.wikipedia.org/wiki/Monty_Hall_problem#Variants https://en.wikipedia.org/wiki/Monty_Hall_problem#Variants
- tzs 7y agoAs others have pointed out, that doesn't quite work because you have not specified that you know where the prize is and make sure to not open that door when opening your 999 998 doors. You can sidestep this by playing a slightly different but equivalent game. The player chooses a door. The host then gives the player the choice of sticking with that door, or instead taking all of the other doors.
- drfuchs 7y agoOf course, that’s the definition of Monty’s Paradox, and exactly how ”Let’s Make A Deal” worked: Monty knows where the one good prize is, and always shows you only bad doors. Anything else is indeed a different story, and a different problem.
- JadeNB 7y agoI'm a mathematician, but I've always wondered when this silver bullet is suggested: am I the only person for whom this exaggerated strategy offers no additional explanatory value? I know, and can prove mathematically, the right answer, but it's no more intuitive to me with 1000000 doors than with 3.
- _asummers 7y agoThe missing part of the explanation (imo), at least the part that helps me explain it in my own brain, is said but not explicit. Imagine the actual mechanics of Monty opening the doors, one by one, and when he gets to that 223,205 door as stated, he skips it, and keeps skipping it. Why _that_ door? What makes it special? I don't think expanding the number of doors without that bit of intuition gives you much.
- svara 7y agoThat's interesting - for me it was the first time seeing it explained that way, and it immediately made perfect sense. I think the intuition behind is that, when you initially choose one door out of 1000000, you are almost sure not to have picked the right one, so you ignore the fact that you even chose one. Then, the host opens 999998 doors, but leaves one closed. That door is now really special! Or a different way to look at it: The host gives you partial information about what's between the 999999 doors you didn't choose. Doesn't it seem really unlikely that your intial choice matters given that there are so many doors?
- crummy 7y agoI feel the same way. My gut says, now there are only two closed doors - it's still a 50/50 chance! This explanation offered above makes sense to my "gut" however: https://news.ycombinator.com/item?id=21349295 https://news.ycombinator.com/item?id=21349295
- JadeNB 7y agoYeah, I agree—that was my first time seeing that explanation, and it really helped me to feel the intuition.
- 7y ago
- hinkley 7y agoYou want to have an ugly argument with other developers, bring up the Monty Hall problem. People who are wrong will not let it go. We had it beaten into us not to use truth tables to solve problems in college, because it doesn't scale. But the Monty Hall problem has a finite set of states you can easily write on a piece of paper. And then you'll see that it jumps from 2/6 successes to 3/6. I tried to replicate that table under duress once (see initial comment) and I couldn't get it sorted. I should try it again and write it down this time. Edit: Why do work when you can find someone else who already has: https://www.statisticshowto.datasciencecentral.com/probability-and-statistics/monty-hall-problem/ https://www.statisticshowto.datasciencecentral.com/probabili... and particular to my statement, this table: https://www.statisticshowto.datasciencecentral.com/wp-content/uploads/2014/12/monty-hall-problem2.png https://www.statisticshowto.datasciencecentral.com/wp-conten... When switching the number of lose scenarios is lower than a naive expectation of outcomes.
- nothrabannosir 7y agoDon’t have the debate; just break out a deck of cards and simulate 30ish rounds. Should be done in 10 minutes, or however small p needs to become for the winds to shift. Then, answer questions.
- hinkley 7y agoYou damn kids and your Monte Carlo’s. When I was a kid we did everything deterministically and we were grateful! And we didn’t have no fancy tablets either. You wanted a mobile computer, you had to steal a cart from Larry. And Larry didn’t like people stealin’ his carts, no sir.
- tantalor 7y ago> We had it beaten into us not to use truth tables to solve problems in college Wait, really? When did you go to college? In mid 2000s, we used truth tables every day studying boolean algebra, logic gates, circuits, etc.
- hinkley 7y ago
- ksaj 7y agoThat's how I prove it with playing cards. Except I use a handful of cards, and not a million of them. The trick is in the wording. With 3 you are literally saying "let me remove all the cards except for a random one, the one you chose, and the one I chose, of which, one is the correct card to choose. Do you want to change your choice?" So it plays out as expected with 3, and the obviousness of the solution when done with 52 cards.
- Myrmornis 7y agoThat might make people think they've understood, but I question whether it is obviously instructive regarding the 3 door problem.
- patrickthebold 7y agoI'll throw in my explanation too: Your first choice is either right or it's wrong. If you switch, right becomes wrong and wrong becomes right. Your first choice is 1/3 right 2/3 wrong.
- bhrgunatha 7y agoThe brilliance of the problem is that you're led to think that the odds have changed. I think there's a psychological component too. So many probability questions are about random events so you (very naturally) take a position that the revealed door is randomly chosen. It is NOT - you always see one of the goats.
- acomjean 7y ago> I think there's a psychological component too. You may be on to something. Perhaps part of it is not wanting to switch from a winning position to a losing one. Thats harder to take so inertia sets in.
- longerthoughts 7y ago>Perhaps part of it is not wanting to switch from a winning position to a losing one This is reminiscent of the endowment effect: https://en.wikipedia.org/wiki/Endowment_effect https://en.wikipedia.org/wiki/Endowment_effect
- WA 7y agoYeah exactly. The Monty Hall problem is about the probability of being correct with your first pick. It doesn’t really have anything to do with "changing probabilities".
- scooble 7y agoAh, that helps. I struggled with this because I didn't see the first choice as relevant to the final outcome. It is easy to see the game as restarted somehow when the 50/50 choice is presented.
- ssalka 7y agoThat really makes it seem like you have more of a 1/2 chance than 2/3.
- sixothree 7y agoIt was supposed to show you your changes had gone from 1/1,000,000 to at least 1/2.
- BrandoElFollito 7y agoBy how does that swap changes your chances? (numerically, that is it goes from 1/100,000,000 to...?)
- mlazos 7y agoThe swap doesn’t change your chances, it’s the host opening one of two other doors that changes your chances. Because the host can open one of two other doors, there are two configurations in which swapping will result in a win but one where not swapping will win.
- rags2riches 7y agoThe host opening the door before the swap is just for show. Ignore that part. The swap is really just picking the two other doors instead of the one you picked first.
- rags2riches 7y agoThe swap is the same as picking all the doors you didn't pick at first. The probability of winning when swapping is the complement of the probability of you winning on your initial pick. There is no third outcome.
- enriquto 7y agoThis is the famous Feynman method for gaining intuition of many physical problems: just set one of the variables to an absurdly huge number and see what happens. He illustrated it with the problem of what happens to the level of water on a pool when you throw overboard a stone from a floating boat. Does the water level rise or drop? Well, imagine that your small stone is made of neutron star material...
- deanCommie 7y agoI'm actually not finding this analogy helpful. Forgive me if I'm missing something obvious, I have a newborn and am underslept. I assume neutron star material is incredibly dense and so the rock would be unbelievably heavy. To the point that the mass would actually likely sink the boat, and then we can't throw the rock out. OK but let's assume that the boat is somehow massive enough to support the weight of the rock. Now we've changed the scale of the other variables of the experiment in proportion to the rock/neutron star, and we're back to ground zero, are we not?
- enriquto 7y agoA 1-ton neutron-star stone will be the size of a dust speck. When it is aboard, only its mass matters, and it will push the boat downwards and displace 1 ton of water. When it is at the bottom of the pool, only its volume matters, and it will displace a negligible amount of water.
- wdr1 7y agoThat didn't work for me. Because if you had been choosing randomly (i.e., did not know which door had the prize), my understand is that the last two doors are still 50/50? It's only when the person opening the door knows where the prize is, that I should always switch?